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Unformatted text preview: total feed rate of Feed 2 =
50/0.6 = 83.3 kmol/h. The total feed rate of A = 50 + 50 = 100 kmol/h. For a recovery of 95%
of A in the distillate, the flow rate of A in the distillate = 0.95(100) = 95 kmol/h. With a mole
fraction of 0.95 for A in the distillate, the total flow rate of the distillate = 95/0.95 = 100 kmol/h.
From Eq. (73) for α = 3, the equilibrium mole fractions of A are related by,
3x
αx
(1)
y=
=
1 + (α − 1) x 1 + 2 x
Equation (1) is plotted in the McCabeThiele diagram on the next page. Because Feed 2 is richer
in A than Feed 1, Feed 2 enters the column above Feed 1. At minimum reflux, the pinch
condition will occur at either Feed 1 or Feed 2. Assume that the pinch occurs at Feed 2. For a
saturated liquid feed, using Eq. (1), the upper section operating line will intersect the equilibrium
curve for xF = 0.6 at y = 3(0.6)/[1 + 2(0.6)] = 0.818. Therefore, the slope of this operating line is,
(L/V)min = (0.95  0.818)/(0.950  0.6) = 0.377
Correspondingly, using Eq. (717), R = L/D = (L/V)min/[1  (L/V)min] = 0.377/(1  0.377) = 0.605
and Lmin = 0.605(100) = 60.5 kmol/h.
Now check the middle section to see if the operating line there is below the equilibrium curve.
The liquid rate in the middle section = L' = L + F2 = 60.5 + 83.3 = 143.8 kmol/h.
The vapor rate in the middle section = V' = V = L + D = 60.5 + 100 = 160.5 kmol/h.
Therefore the slope of the operating line in the middle section = L'/V' = 143.8/160.5 = 0.896.
As seen in the McCabeThiele diagram on the next page, this operating line does not cross over
the equilibrium curve. Therefore, the pinch does occur at Feed 2 (the upper feed).
For an operating reflux ratio of 1.33 times minimum, L = 1.33(60.5) = 80.5 kmol/h.
The vapor rate in the upper section = L + D = 80.5 + 100 = 180.5 kmol/h. Analysis: (continued) Exercise 7.30 (continued) Therefore the upper section operating line has a slope, L/V = 80.5/180.5 = 0.446 and passes
through the point y = x = 0.95. It intersects the vertical qline at x = 0.6 and for the slope of
0.446 = (0.95  y)/(0.95  0.6), y = 0.794.
For the middle section, L' = L + F2 = 80.5 + 83.3 = 163.8 kmol/h and V' = V = 180.5 kmol/h
Therefore, the middle section operating line has a slope of L'/V' = 163.8/180.5 = 0.908 and
intersects the qline for x = 0.6 at y = 0.794. It intersects the vertical qline at x = 0.4 and for the
slope of 0.908 = (0.794  y)/(0.6  0.4), y = 0.613.
For the lower section, L" = L' + F1 = 163.8 + 125 = 288.8 kmol/h and V"=V' = 180.5 kmol/h
Therefore, the lower section operating line has a slope of L"/V" = 288.8/180.5 = 1.60 and
intersects the qline for x = 0.4 and y = 0.613. As seen in Fig. 7.27(c), the mole fraction of A in
the bottoms, xB , is determined from the intersection of the operating line for the lower section
with the yaxis. Thus, 1.60 = (0.613  0)/(0.4  xB). Solving, xB = 0.0169 for component A.
Since the bottoms contains 5 kmol/h of A, the bottoms rate = B = 5/0.0169 = 295.9 kmol/h.
Thus, t...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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